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 Post subject: GMAT Number Theory
PostPosted: Tue Aug 04, 2009 1:26 pm 
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Joined: Sun Apr 19, 2009 6:56 pm
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When 777 is divided by positive integer n, the remainder is 77. How many possibilities are there for n?


How can we assume that since there are 5 factors less than or equal to 7: 1 , 2 , 4 , 5 , and 7, there are 5 possible values of n (i.e. factors of 700) greater than 77 ? Am I missing some link?


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 Post subject: Re: Test 5, question 10
PostPosted: Tue Aug 04, 2009 1:31 pm 
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If the remainder is 77, then n must logically be greater than 77. Also, there must be a positive integer q such that 777= nq + 77. i.e. nq = 700. Therefore, the factors of 700 greater than 77 comprise the possible values of n. Instead of counting the factors of 700 that are greater than 77, let’s count the ones that are less than or equal to 700/77.As 700 = 50 × 2 × 7, we can see that there are 5 factors of 700 that are less than or equal to 7: 1 , 2 , 4 , 5 , and 7. Thus there are 5 possible values of n (i.e. factors of 700) greater than 77. They are 700, 350, 175, 140 and 100. A simpler way to think about this is to figure out all the divisors of 700 bigger than 77. That gives you 700, 350, 175, 140 and 100.


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 Post subject: Re: Test 5, question 10
PostPosted: Wed Mar 02, 2011 3:17 am 
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Why do we count the factors that are less than or equal to 700/77 ?


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 Post subject: Re: Test 5, question 10
PostPosted: Wed Mar 02, 2011 5:45 am 
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Imagine, that we have a factor of 700, a, which is greater than 77. Then we know that:
a > 77
700 = ab, so a = 700/b.
Therefore 700/b > 77 or 700/77 > b

As you see, for each factor a, which is greater than 77, there is a corresponding factor b = 700/a, which is less than 700/77.

Furthermore, it's a very good property to remember. It helps to analyze if an integers is a prime number or to find all the possible factors. For example:

How many distinct factors does 127 have?

It's easy to see that 11 < √127 < 12.
As we already know from the above reasoning, any factor a < √127 will yield a factor b > 127/√127 = √127. So we have to check divisibility by only 11 numbers and that will give us all the possible divisors.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11

1) 127 = 1 × 127 is obvious. So we have two factors here.
2) 127 is an odd number. So it's not divisible by any even number.

This leaves us:
3, 5, 7 , 9, 11

3) 127 is NOT divisible by 3, since 1 + 2 + 7 = 11 is NOT.
4) 127 is NOT divisible by 5, since it doesn't end in 0 or 5.

This leaves us:
7, 11

5) 127 = 121 + 6 = 11² + 6, so it's NOT divisible by 11
6) 127/7 = 16 × 7 + 1 (you can do it by long division method), so it's NOT divisible by 7.

As the result, we know that 127 is a prime number and has only two distinct factors: 1, 127.


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